Post on 16-Oct-2021
HW3 Physics 311 Mechanics
Fall 2015Physics department
University of Wisconsin, Madison
Instructor: Professor Stefan Westerhoff
By
Nasser M. Abbasi
November 28, 2019
Contents
0.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70.4 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90.5 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
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3
0.1 Problem 1
Mechanics
Physics 311Fall 2015
Homework 3 (9/25/15, due 10/2/15)
1. (5 points)A uniform rope of total mass m and total length l lies on a table, with a length z hangingover the edge. Find the differential equation of motion.
2. (10 points)A particle of mass m perched on top of a smooth hemisphere of radius R is disturbedslightly, so that it begins to slide down the side. Use Lagrange multipliers to find thenormal force of constraint exerted by the hemisphere on the particle and determine theangle relative to the vertical at which it leaves the hemisphere.
3. (10 points)Consider the object shown in the figure below, which has a half-sphere of radius a as thebottom part and a cone on top. The center of mass (P ) is at a distance b from the groundwhen the object is standing upright. Let I be the moment of inertia. Find the frequencyof small oscillations if the object is disturbed slightly from its upright position. Whathappens if a = b or b > a?
...continued on next page...
SOLUTION
z
l β z
U = 0
U = β 12z
zlmg
T = 12zlmz2 + 1
2lβzl mz2
C.M. at half way
The top portion of the rope moves with same speed as the hanging portion. Hence π§ is usedto describe the motion as the generalized coordinate. From the above
π = β οΏ½12π§οΏ½ οΏ½
π§ποΏ½ππ = β
12 οΏ½π§2
π οΏ½ππ
π =12οΏ½π§ποΏ½ποΏ½οΏ½2 +
12 οΏ½π β π§π οΏ½ποΏ½οΏ½2 =
12ποΏ½οΏ½2
In finding π we used 12 since the center of mass of the hanging part is half way over the
length. So the potential energy is taken from the center of mass. In the above, οΏ½οΏ½ is usedfor both parts of the rope, since both parts move with same speed. Applying Lagrangianequations gives
πΏ = π β π
=12ποΏ½οΏ½2 +
12 οΏ½π§2
π οΏ½ππ
HenceππΏππ§
=π§πππ
πππ‘ππΏποΏ½οΏ½
= ποΏ½οΏ½
4
And thereforeπππ‘ππΏποΏ½οΏ½
βππΏππ§
= 0
ποΏ½οΏ½ βπ§πππ = 0
οΏ½οΏ½ =π§ππ
When π§ = 0 then the acceleration is zero as expected. When π§ = π2 then οΏ½οΏ½ = 1
2π and whenπ§ = π then οΏ½οΏ½ = π as expected since in this case the rope will all be falling down on its ownweight due to gravity and should have π as the acceleration.
0.2 Problem 2
Mechanics
Physics 311Fall 2015
Homework 3 (9/25/15, due 10/2/15)
1. (5 points)A uniform rope of total mass m and total length l lies on a table, with a length z hangingover the edge. Find the differential equation of motion.
2. (10 points)A particle of mass m perched on top of a smooth hemisphere of radius R is disturbedslightly, so that it begins to slide down the side. Use Lagrange multipliers to find thenormal force of constraint exerted by the hemisphere on the particle and determine theangle relative to the vertical at which it leaves the hemisphere.
3. (10 points)Consider the object shown in the figure below, which has a half-sphere of radius a as thebottom part and a cone on top. The center of mass (P ) is at a distance b from the groundwhen the object is standing upright. Let I be the moment of inertia. Find the frequencyof small oscillations if the object is disturbed slightly from its upright position. Whathappens if a = b or b > a?
...continued on next page...
SOLUTION
ΞΈ
polar position (r, ΞΈ)
U = mgr sin ΞΈ
g
R
T = 12m(r2 + r2ΞΈ2)
constraint f(r, ΞΈ) = r βR = 0
Generalized coordinates used r, ΞΈ
There are two coordinates π, π (polar) and one constraint
π (π, π) = π β π = 0 (1)
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Now we set up the equations of motion for π
π =12π οΏ½οΏ½οΏ½2 + π2οΏ½οΏ½2οΏ½
π = πππ sinππΏ = π β π
=12π οΏ½οΏ½οΏ½2 + π2οΏ½οΏ½2οΏ½ β πππ sinπ
Hence the Euler-Lagrangian equations are
πππ‘ππΏποΏ½οΏ½
βππΏππ
+ πππππ
= 0 (2)
πππ‘ππΏποΏ½οΏ½
βππΏππ
+ πππππ
= 0 (3)
Butπππ‘ππΏποΏ½οΏ½
= ποΏ½οΏ½
ππΏποΏ½οΏ½
= ππ2οΏ½οΏ½
πππ‘ οΏ½
ππΏποΏ½οΏ½οΏ½
= π οΏ½2ποΏ½οΏ½οΏ½οΏ½ + π2οΏ½οΏ½οΏ½
ππΏππ
= πποΏ½οΏ½2 β ππ sinπ
ππΏππ
= βπππ cosπ
ππππ
= 1
ππππ
= 0
Hence (2) becomes
ποΏ½οΏ½ β πποΏ½οΏ½2 + ππ sinπ + π = 0 (4)
And (3) becomes
ποΏ½2ποΏ½οΏ½οΏ½οΏ½ + π2οΏ½οΏ½οΏ½ + πππ cosπ = 0ποΏ½οΏ½ + 2οΏ½οΏ½οΏ½οΏ½ + π cosπ = 0 (5)
We now need to solve (1,4,5) for π. Now we have to apply the constrain that π = π in theabove to be able to solve (4,5) equations. Therefore, (4,5) becomes
βππ οΏ½οΏ½2 + ππ cosπ + π = 0 (4A)
π οΏ½οΏ½ + π cosπ = 0 (5A)
Where (4A,5A) were obtained from (4,5) by replacing π = π and οΏ½οΏ½ = 0 and οΏ½οΏ½ = 0 since weare using that π = π which is constant (the radius).
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From (5A) we see that this can be integrated giving
π οΏ½οΏ½2 + 2π sinπ + π = 0 (6)
Where π is constant. Since if we diοΏ½erentiate the above with time, we obtain
2π οΏ½οΏ½οΏ½οΏ½ + 2ποΏ½οΏ½ cosπ = 0π οΏ½οΏ½ + π cosπ = 0
Which is the same as (5A). Therefore from (6) we find οΏ½οΏ½2 to use in (4A). Hence from (6)
οΏ½οΏ½2 = β2ππ
sinπ + π
To find π we use initial conditions. At π‘ = 0, π = 900 and οΏ½οΏ½ (0) = 0 hence
π = 2ππ
Therefore
οΏ½οΏ½2 = β2ππ
sinπ + 2 ππ
= 2ππ (1 β sinπ)
Plugging the above into (4A) in order to find π gives
βππ οΏ½2ππ (1 β sinπ)οΏ½ + ππ sinπ + π = 0
π = π οΏ½2π (1 β sinπ)οΏ½ β ππ sinππ = 2ππ β 2ππ sinπ β ππ sinπ= ππ (2 β 3 sinπ)
Now that we found π ,we can find the constraint force in the radial direction
π = πππππ
= ππ (2 β 3 sinπ)The particle will leave when π = 0 which will happen when
2 β 3 sinπ = 0
π = sinβ1 οΏ½23οΏ½
= 41.80
Therefore, the angle from the vertical is
90 β 41.8 = 48.20
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g
ΞΈ0 = 41.8048.20
N = Ξ»βfβr
N
Particle will leave when N = 0
v
constraint force
0.3 Problem 3
Mechanics
Physics 311Fall 2015
Homework 3 (9/25/15, due 10/2/15)
1. (5 points)A uniform rope of total mass m and total length l lies on a table, with a length z hangingover the edge. Find the differential equation of motion.
2. (10 points)A particle of mass m perched on top of a smooth hemisphere of radius R is disturbedslightly, so that it begins to slide down the side. Use Lagrange multipliers to find thenormal force of constraint exerted by the hemisphere on the particle and determine theangle relative to the vertical at which it leaves the hemisphere.
3. (10 points)Consider the object shown in the figure below, which has a half-sphere of radius a as thebottom part and a cone on top. The center of mass (P ) is at a distance b from the groundwhen the object is standing upright. Let I be the moment of inertia. Find the frequencyof small oscillations if the object is disturbed slightly from its upright position. Whathappens if a = b or b > a?
...continued on next page...
SOLUTION
8
a c.m.
bb
hΞΈ
h = aβ (aβ b) cos ΞΈ
a
U = mgh = mg(aβ (aβ b) cos ΞΈ
From the above, we see that the center of mass has height above the ground level afterrotation of
β = π β (π β π) cosπTaking the ground state as the floor, the potential energy in this state is
π = ππβ= ππ (π β (π β π) cosπ)
And the kinetic energy
π =12πΌοΏ½οΏ½2
Hence the Lagrangian is
πΏ = π β π
=12πΌοΏ½οΏ½2 β ππ (π β (π β π) cosπ)
Therefore the equation of motion is
πππ‘ππΏποΏ½οΏ½
βππΏππ
= 0
πΌοΏ½οΏ½ βπππ οΏ½
12πΌοΏ½οΏ½2 β ππ (π β (π β π) cosπ)οΏ½ = 0
πΌοΏ½οΏ½ +πππππ (π β (π β π) cosπ) = 0
πΌοΏ½οΏ½ βπππππ (π β π) cosπ = 0
πΌοΏ½οΏ½ + ππ (π β π) sinπ = 0For small π, sinπ β π, hence the above becomes
οΏ½οΏ½ +ππ (π β π)
πΌπ = 0
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Therefore the natural angular frequency is
ππ = οΏ½ππ(πβπ)
πΌ
When π = π then ππ = 0 and the mass do not oscillate but remain at the new positions. Whenπ > π then ππ is complex valued. This is not possible, as the natural frequency must be real.So center of mass can not be in the upper half.
0.4 Problem 4
4. (15 points)A sphere of radius r, mass m, and moment of inertia I = 2
5mr
2 is contrained to roll withoutslipping on the lower half of the inner surface of a hollow cylinder of inside radius R (whichdoes not move). Let the z-direction go along the axis of the cylinder.(1) Determine the Lagrangian, the equations of motion, and the period for small oscilla-tions. Ignore a possible motion in the z-direction.(2) Determine the Lagrangian in the more general case where the motion in the z-directionis included. Describe the motion in the z-direction.
5. (10 points)Consider a disc of mass m and radius a that has a string wrapped around it with oneend attached to a fixed support and allowed to fall with the string unwinding as it falls.(This is essentially a yo-yo with the string attached to a finger held motionless as a fixedsupport.) Find the equation of motion of the disc.
SOLUTION
10
r
R
ΞΈ
Ο
Ο(Rβ r)ΞΈ
rΟ
No slip condition
(Rβ r)ΞΈ = rΟ x
y
2 generalized coordinates ΞΈ, Ο butconstraint reduces this to one coor-dinate ΞΈ
h = Rβ (Rβ r) cos ΞΈ
Part (1): There are two coordinates are π, π, but due to dependency between them (no slip)then this reduces the degree of freedom by one, and there is one generalized coordinate π.The constraints of no slip means
π οΏ½π, ποΏ½ = (π β π) π β ππ = 0
Which means the center of the small disk move in speed the same as the point of the diskthat moves on the edge of the larger cylinder as shown in the figure above.
π =12πΌοΏ½οΏ½2 +
12π οΏ½(π β π) οΏ½οΏ½οΏ½
2
π = ππβ = ππ (π β (π β π) cosπ)
Using πΌ = 25ππ
2 and using οΏ½οΏ½ = (π βπ)π οΏ½οΏ½ from the constraint conditions, then π becomes
π =12 οΏ½25ππ2οΏ½ οΏ½
(π β π)π
οΏ½οΏ½οΏ½2
+12π οΏ½(π β π) οΏ½οΏ½οΏ½
2
=15π (π β π)2 οΏ½οΏ½2 +
12π (π β π)2 οΏ½οΏ½2
=710π (π β π)2 οΏ½οΏ½2
Hence
πΏ = π β π
=710π (π β π)2 οΏ½οΏ½2 β ππ (π β (π β π) cosπ)
AndππΏππ
= βππ (π β π) sinπ
ππΏποΏ½οΏ½
=75π (π β π)2 οΏ½οΏ½
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Therefore the equation of motion is
πππ‘ππΏποΏ½οΏ½
βππΏππ
= 0
75π (π β π)2 οΏ½οΏ½ + ππ (π β π) sinπ = 0
οΏ½οΏ½ +π
75(π β π)
sinπ = 0
For small angle
οΏ½οΏ½ +5π
7 (π β π)π = 0
The frequency of oscillation is
ππ =οΏ½
5π7 (π β π)
Using ππ =2ππ then the period of oscillation is
π =2π
οΏ½5π
7(π βπ)
= 2ποΏ½
7 (π β π)5π
Part (2):
There are now two generalized coordinates, π and π§. The sphere now rotates in 2 angularmotions, οΏ½οΏ½ which is the same as it did in part 1, and in addition, it rotate with angularmotion, οΏ½οΏ½ which is rolling down the π§ axis. The new constraint is that
π1 (πΌ, π§) = π§ β ππΌ = 0 (1)
So that no slip occurs in the π§ direction. This is in additional of the original no slip conditionwhich is
π2 οΏ½π, ποΏ½ = (π β π) π β ππ = 0 (2)
The following diagram illustrates this
z
The sphere is now distance zaway from the origin. Thereis new constraint now asshown
z axis
RΞ±
z = rΞ±
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Now there are translation kinetic energy in the π§ direction as well as new rotational kineticenergy due to spin πΌ. Therefore
π =
part(1)
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½12πΌοΏ½οΏ½2 +
12π οΏ½(π β π) οΏ½οΏ½οΏ½
2+
due to moving in z
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½12ποΏ½οΏ½2 +
12πΌοΏ½οΏ½2
π = ππβ = ππ (π β (π β π) cosπ)Notice that the potential energy do not change, since it depends only on the height abovethe ground. Using πΌ = 2
5ππ2 and from constraints (1,2) then π becomes
π =12 οΏ½25ππ2οΏ½
οΏ½οΏ½2
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½(π β π)π
οΏ½οΏ½οΏ½2
+12π οΏ½(π β π) οΏ½οΏ½οΏ½
2+12ποΏ½οΏ½2 +
12 οΏ½25ππ2οΏ½
οΏ½οΏ½2οΏ½οΏ½οΏ½οΏ½ποΏ½2
= οΏ½15ππ2οΏ½
(π β π)π2
οΏ½οΏ½2 +12π (π β π)2 οΏ½οΏ½2 +
12ποΏ½οΏ½2 + οΏ½
15ππ2οΏ½
οΏ½οΏ½2
π2
=710π (π β π) οΏ½οΏ½2 +
710ποΏ½οΏ½2
Hence the Lagrangian is
πΏ = π β π
=710π (π β π) οΏ½οΏ½2 +
710ποΏ½οΏ½2 β ππ (π β (π β π) cosπ)
This part only now asks for motion in π§ direction. Hence
ππΏππ§
= 0
ππΏποΏ½οΏ½
=75ποΏ½οΏ½
Since ππΏππ§ = 0 then
πππ‘ππΏποΏ½οΏ½
= 0
Hence ππΏποΏ½οΏ½ is the integral of motion. Or
75ποΏ½οΏ½ = 0
or
οΏ½οΏ½ = 0οΏ½οΏ½ = π
Where π is constant. This means the sphere rolls down the π§ axis at constant speed.
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0.5 Problem 5
4. (15 points)A sphere of radius r, mass m, and moment of inertia I = 2
5mr
2 is contrained to roll withoutslipping on the lower half of the inner surface of a hollow cylinder of inside radius R (whichdoes not move). Let the z-direction go along the axis of the cylinder.(1) Determine the Lagrangian, the equations of motion, and the period for small oscilla-tions. Ignore a possible motion in the z-direction.(2) Determine the Lagrangian in the more general case where the motion in the z-directionis included. Describe the motion in the z-direction.
5. (10 points)Consider a disc of mass m and radius a that has a string wrapped around it with oneend attached to a fixed support and allowed to fall with the string unwinding as it falls.(This is essentially a yo-yo with the string attached to a finger held motionless as a fixedsupport.) Find the equation of motion of the disc.
SOLUTION
This is first solved using energy method, then solved using Newton method.
T
aΞΈ
g
constraint: ya = ΞΈ
y
Energy method
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Constraint is π οΏ½π¦, ποΏ½ = π¦ β ππ = 0. Hence οΏ½οΏ½ = οΏ½οΏ½π
π = βπππ¦
π =12πΌοΏ½οΏ½2 +
12ποΏ½οΏ½2
=12πΌ οΏ½οΏ½οΏ½ποΏ½2+12ποΏ½οΏ½2
=12 οΏ½
12ππ2οΏ½ οΏ½
οΏ½οΏ½ποΏ½2+12ποΏ½οΏ½2
=14ποΏ½οΏ½2 +
12ποΏ½οΏ½2
=34ποΏ½οΏ½2
Hence
πΏ = π β π
=34ποΏ½οΏ½2 + πππ¦
ThereforeππΏππ¦
= ππ
ππΏποΏ½οΏ½
=32ποΏ½οΏ½
πππ‘ππΏποΏ½οΏ½
=32ποΏ½οΏ½
And the equation of motion becomes
πππ‘ππΏποΏ½οΏ½
βππΏππ¦
= 0
32ποΏ½οΏ½ β ππ = 0
οΏ½οΏ½ =23π
Newton method
Using Newton method, this can be solved as follows. The linear equation of motion is(positive is taken downwards)
πΉ = ποΏ½οΏ½βπ + ππ = ποΏ½οΏ½ (1)
And the angular equation of motion is given by
ππ = πΌοΏ½οΏ½ (2)
15
Due to constraint π οΏ½π¦, ποΏ½ = π¦ β ππ = 0, thenοΏ½οΏ½π= οΏ½οΏ½
Using the above in (2) gives
ππ = πΌοΏ½οΏ½π
π = πΌοΏ½οΏ½π2
(3)
Replacing π in (1) with the π found in (3) results in
ποΏ½οΏ½ = βπΌοΏ½οΏ½π2+ ππ
οΏ½οΏ½ οΏ½π +πΌπ2 οΏ½
= ππ
οΏ½οΏ½ =ππ
π + πΌπ2
But πΌ = 12ππ
2 then the above becomes
οΏ½οΏ½ =ππ
π +12ππ
2
π2
=π
1 + 12
=23π
Which is the same (as would be expected) using the energy method